The median of this grouped distribution must be interpolated first, and the mean deviation is then measured from it using class marks.

Median class $=20$–$30$ (c.f. $=35$ first reaches 25). $M=20+((25-15)/20)\times10=25$. …

Step 1 — Recall the cumulative frequency table (as in Worked Example 3).

Step 2 — Locate the median class and interpolate. $N=50$, $N/2=25$. The c.f. first reaches or exceeds 25 at class 20–30 (c.f.$=35$): $L=20$, $cf=15$, $f=20$, $h=10$.

Step 3 — Find $|x-25|$ and $f|x-25|$ using class marks.

| Class | $f$ | Class mark $x$ | $|x-25|$ | $f|x-25|$ |

|---|---|---|---|---|

| 0–10 | 5 | 5 | 20 | 100 |

| 10–20 | 10 | 15 | 10 | 100 |

| 20–30 | 20 | 25 | 0 | 0 |

| 30–40 | 10 | 35 | 10 | 100 |

| 40–50 | 5 | 45 | 20 | 100 |

| Total | 50 | | | 400 |

Step 4 — Divide, then find the coefficient.

$\text{MD}(M) = \dfrac{400}{50} = 8$ …

Using the mean (found by a different method) in place of the median here would silently answer a different question — this example specifically asks for deviations about the MEDIAN, located by interpolatio …